After reduction of a problem, I find myself in front of these integrals, $$ \int_0^{2\pi} cn \left( \frac{2K(1/2)}{\pi} \theta, \frac{1}{2} \right) \cos(m \theta) \mathrm{d}\theta, $$ with $m$ positive integers. For now,

  • I think it vanishes for $m$ even,
  • I am not sure how to get the value for $m=1$,
  • I am also not sure whether an expansion of $\cos(m\theta)$ then integration by parts would yield something nice (in terms of the case $m=1$ hopefully).

I would like to hear your thoughts on these ones!

  • 2
    $\begingroup$ $\S 22.11$ of DLMF has several formula for the Fourier series of Jacobi elliptic functions, $\endgroup$ Apr 26, 2019 at 20:11
  • $\begingroup$ Thank you, I'm looking into it, that might be helpful! $\endgroup$
    – Cryme
    Apr 26, 2019 at 20:14
  • $\begingroup$ It worked out, I posted the answer for anyone's interest $\endgroup$
    – Cryme
    Apr 26, 2019 at 20:35

1 Answer 1


As pointed by @achille hui, the Fourier series of Jacobi elliptic functions are known. In particular here, $$ cn \left( \frac{2K(1/2)}{\pi} \theta, \frac{1}{2} \right) = \frac{2\sqrt{2}\pi}{K(1/2)} \sum_{n=0}^\infty \frac{e^{-(n+1/2)\pi}}{1+e^{-(2n+1)\pi}} \cos ((2n+1) \theta), $$ in turn, $$ \int_0^{2\pi} cn \left( \frac{2K}{\pi} \theta, \frac{1}{2} \right) \cos(m \theta) \mathrm{d}\theta = \begin{cases} \frac{2\sqrt{2}}{K(1/2)} \frac{e^{-m\pi/2}}{1+e^{-m\pi}} \pi^2 &\textrm{if $m$ is odd}\\ 0 &\textrm{otherwise} \end{cases} $$


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